Question

Using vectors, find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)

Comments

It’s entirely formula based. Why don’t you try it on your own instead of asking?

Answer 1

Given three points A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5) forming a triangle. Let position vectors of the vertices A, B and C of ΔABC be \(\vec a,\,\vec b\) and  \(\vec c\) respectively.

We know position vector of a point (x, y, z) is given by  \(\text x\hat i+y\hat j+z\hat k\),  where \(\hat i\), \(\hat j\)  and \(\hat k\) are unit vectors along X, Y and Z directions.

To find area of ΔABC, we need to find at least two sides of the triangle. So, we will find vectors \(\vec{AB}\) and \(\vec{AC}\).

Recall the vector \(\vec{AB}\) is given by

\(\vec{AB}\) = position vector of B - position vector of A

Similarly, the vector \(\vec{AC}\) is given by

\(\vec{AC}\) = position vector of C - position vector of A

Recall the area of the triangle whose adjacent sides are given by the two vectors

Here, we have (a₁, a₂, a₃) = (1, 2, 3) and (b₁, b₂, b₃) = (0, 4, 3)

Thus, area of the triangle is \(\cfrac{\sqrt{61}}2\) square units.